The Monty Hall Problem Solved, Once and For All

The first time I ever encountered the infamous Monty Hall Problem was 5 years ago on a video from one of my favorite YouTubers, Numberphile. The problem states:

Source: Wikipedia

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to switch your choice?"

The answer is a resounding yes, you are twice as likely to win by switching. However, the problem is seemingly paradoxical and to this day baffles many people - why does switching doors matter in the grand scheme? Through hundreds of YouTube comments on the original videos, I see 2 major misconceptions that I will layout and explain the best that I can. 

Misconception 1: The event of the door not opened/picked is independent of the door the host picked, thus the probability of winning by switching is $\frac{1}{2}$

This thought process is simply a misunderstanding of independence, which in of itself is a very unintuitive topic. We often think of the idea as synonymous to no correlation or association, but that definition doesn't quite fit. Independence implies no correlation but the converse is not true - 0 correlation does not imply independence! Thus, rather than an intuitive explanation, independence is best explained by the following formula: 

Events $A$ and $B$ are independent $\iff \mathbb{P}[A \wedge B] = \mathbb{P}[A] \cdot \mathbb{P}[B]$.

For any two events, $\mathbb{P}[A \wedge B] = \mathbb{P}[A | B] \cdot \mathbb{P}[B]$ and vice versa, (switching $A$ and $B$), so we can conclude if two events are independent, $\mathbb{P}[A | B] = \mathbb{P}[A]$

That's a lot of mathy-math terms to make me sound smart and credible. In English, this says that if two events are independent, their joint probability is the same as the product of their probability separately, and that the probability of a first event given that a second event is true is the same as the probability of the first event (intuitively, independence says one event happening shouldn't affect the probability of another!) Let's lay out an example to see why this is untrue. 

Label the 3 doors as follows: picked, shown, and switched. Let $A$ be the event that the switched door is the prize and $B$ be the event that the shown door is a goat. The probability of $A$ is $\frac{1}{3}$ because only one of the 3 doors are prizes. However, $\mathbb{P}[A | B]$ (Event $A$ given event $B$ is true) is not $\frac{1}{3}$, it is $\frac{1}{2}$ because given door $B$ is given to be a goat! Thus these events are clearly not independent.

Misconception 2: If you list out all the possible outcomes, there are 6 events in which you win and 6 events in which you lose, thus the probability of winning by switching is $\frac{1}{2}$

This claim is true if you only consider the number of possible events, so the claim baffled me for a long time as a high schooler. However, the key insight, which also helps solve the Monty Hall Problem, comes in the PROBABILITY of each event happening - the claim suggests that each of the 12 events as equally likely, which is untrue.

Source: Summer 2019 CS70 Lecture 15 slides by Elizabeth Yang

The diagram shown at the bottom right of the picture lays out all possibilities of the game by using the switching strategy - the first row (labeled $c_1$) contains all the possible outcomes of prize doors (door 1, 2, or 3), each occurring with a probability of $\frac{1}{3}$. The second row (labeled $c_2$) contains all the possible doors the contestant first chooses, again each occurring with a probability of $\frac{1}{3}$.

Here's the kicker - the last row ($c_3$) contains all the possible doors that host reveals that contains a goat. As we can see, if initially we choose the prize door, the host has a $\frac{1}{2}$ chance of revealing either of the goat doors (routes highlighted in yellow) and switching would yield a loss, whereas if we initially choose a goat door, we force the host to reveal the other goat door (routes highlighted in blue) and switching would yield a win! Thus, although the yellow routes and blue routes match in number of events, the probability of each route is different - blue routes happen with a probability of $\frac{1}{3} \cdot \frac{1}{3} \cdot 1 = \frac{1}{9}$ while yellow routes happen with a probability of $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{18}$! There are 6 of each route, so the total probability of yellow routes is $\frac{1}{3}$ and the total probability of blue routes is $\frac{2}{3}$.

Thus, by switching doors every time you play the game, you win (blue route) with twice the probability of losing (yellow route). If the explanation is not convincing enough and you are really set on proving it wrong, try setting up and playing 100 games on your own, multiple times, switching every time - if you manage to consistently lose 50 games or more, call me :)